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eduard
3 years ago
5

According to rounding rules for addition the sum of 27.1, 34.538, and 37.68 is

Chemistry
1 answer:
Katena32 [7]3 years ago
5 0
The answer should be...99.318!
You might be interested in
If the Dry bulb reads 25 degrees Celsius and the wet bulb reads 22
Nitella [24]

Answer:

Relative humidity is low .

Explanation:

The wet bulb reads low temperature because due to low humidity of atmosphere , evaporation of water takes place from the wet bulb which makes the bulb cool and therefore it reads lower temperature . In the process of evaporation , heat equal to latent heat of vaporization is taken from the bulb and it loses temperature.

6 0
3 years ago
The states of matter involved in a milkshake
Anna [14]

Answer:

it is a liquid that is made from the combination of a liquid and a solid.

Explanation:

When these two states of matter are ground and mixed together, they form one state that is liquid

6 0
2 years ago
Read 2 more answers
What is the mass of 2.25 x 10^25 atoms of lead
NemiM [27]

Answer: the amount of a substance that contains 6.02 x 1023 respective particles of that

substance

Avogadro’s number: 6.02 x 1023

Molar Mass: the mass of one mole of an element

CONVERSION FACTORS:

1 mole = 6.02 x 1023 atoms 1 mole = atomic mass (g)

Try:

1. How many atoms are in 6.5 moles of zinc?

6.5 moles 6.02 x 1023 atoms = 3.9 x 1024 atoms

1 mole

2. How many moles of argon are in a sample containing 2.4 x 1024 atoms of argon?

2.4 x 1024 atoms of argon 1 mole = 4.0 mol

6.02 x 1023 atoms

3. How many moles are in 2.5g of lithium?

2.5 grams Li 1 mole = 0.36 mol

6.9 g

4. Find the mass of 4.8moles of iron.

4.8 moles 55.8 g = 267.84 g = 270g

1 mole

MOLE PARTICLES

(ATOM)

MASS

(g)

1 mole = molar

mass

(look it up on

the PT!)

1 mole =

6.02 x 1023

atoms

1 mole = 6.02 x 1023 atoms 1 mole = atomic mass (g)

Two Step Problems:

1. What is the mass of 2.25 x 1025 atoms of lead?

2.25 x 1025 atoms of lead 1 mole 207.2g = 7744.19g = 7740g

6.02 x 1023 atoms 1 mole

2. How many atoms are in 10.0g of gold?

10 g gold 1 mole 6.02 x 1023 atoms = 3.06 x 1022 atoms

197.0g 1 mole

PRACTICE PROBLEMS:

1. How many moles are equal to 625g of copper?

625g of copper 1 mol = 9.77 mol Cu

64 g Cu

2. How many moles of barium are in a sample containing 4.25 x 1026 atoms of barium?

4.25 x 1026 atoms of barium 1 mol = 706 mol

6.02 x 1023 atoms

3. Convert 2.35 moles of carbon to atoms.

2.35 moles 6.02 x 1023 atoms = 1.41 x 1024 atoms

1 mole

4. How many atoms are in 4.0g of potassium?

4.0 g 1 mole 6.02 x 1023 atoms = 6.2 x 1022 atoms

39.1 g 1 mole

1 mole = 6.02 x 1023 atoms 1 mole = atomic mass (g)

5. Convert 9500g of iron to number of atoms in the sample.

9500 g Fe 1 mole 6.02 x 1023 atoms = 1.0 x 1026 atoms

55.8g 1 mole

6. What is the mass of 0.250 moles of aluminum?

0.250 moles 27.0g = 6.75 g Al

1 mole

7. How many grams is equal to 3.48 x 1022 atoms of tin?

3.48 x 1022 atoms 1 mole 118.7g = 6.86 g Sn

6.02 x 1023 atoms 1 mole

8. What is the mass of 4.48 x 1021 atoms of magnesium?

4.48 x 1021 atoms 1 mole 24.3g = 0.181 g Mg or 1.8 x 10-1

6.02 x 1023 atoms 1 mole

9. How many moles is 2.50kg of lead?

2.50kg 1000 g 1 mol = 12.1 mol

1 kg 207.2 g

10.Find the mass, in cg, of 3.25 x 1021 atoms of lithium.

3.25 x 1021 atoms 1 mol 6.9g 100cg = 3.7

Explanation:

5 0
3 years ago
6 Apply fill in the genotypes and phenotypes of the
Nuetrik [128]
The scientific geographic name is usually a great election
6 0
3 years ago
Consider the following unbalanced reaction: P4(s) + F2(g) → PF3(g) What mass of fluorine gas is needed to produce 120. g of PF3
Studentka2010 [4]

Answer:

44.28 grams.

Explanation:

Let us write the balanced reaction:

P_{4}+6F_{2}-->4PF_{3}

As per balanced equation, six moles of fluorine gas will give four moles of PF₃.

The mass of PF₃ required = 120 g

The molar mass of PF₃ = 88g/mol

Moles of PF₃ required =\frac{mass}{molarmass}=\frac{120}{88}=1.364mol

The moles of fluorine gas required = \frac{4X1.364}{6}=0.91

the mass of fluorine gas required = moles X molar mass = 0.91x38 = 34.58g

Now this much mass will be required if the reaction is of 100% yield

But as given that the yield of reaction is only 78.1%

The mass of fluorine required = \frac{massX100}{78.1} =\frac{34.58X100}{78.1} =44.28g

4 0
3 years ago
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